Answer
$$ = {{11\,\sqrt{6}\,\ln \left(2\,x^2+3\right)+14\,\arctan \left({{2\,x }\over{\sqrt{6}}}\right)-10\,\sqrt{6}\,x-12\sqrt{6}\,\ln \left(x-1\right)}\over{20\,\sqrt{6}}} $$
Explanation
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